Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

UNIT-IV
81
Examples
1. The polynomial
Proof
is irreducible over Q, where p is a prime.
( x − 1) f ( x)
= x p − 1 . Put x = y + 1 , then
Where
F
p
HG I i K J =
F
HG I K J
F = y p p p
p + HGI KJ F y
p + HGI KJ −1 y
p−2
+ ........... + y
1 2 p − 1
(1)
, i < p
Not that
and
divides the product
. Hence p divides
Dividing (1) by y, we see that
satisfies the hypothesis of Eisentein criterion and so it is irreducible over Q. Hence f ( x) is irreducible
1 p−2
Fi!
yf p p
( p iy
! − )( ) p = ( 2y
).............( x + 1)
........... − 1 p − i
x−
+ 1
)
HG I 2.
i K J F p p
y
y
p
+ ) = + ............
i!
H GI K J F
+
HG I is irreducible over Q, since p = 5 ,
5 4 3
5 2 ×
3 n
( S fy ( = + ×
x
y{ )
20 1 + a= 4 3 2
) 1 ,)
3x a+ 2x (........... −y p+ p − K J
−15 1) xa + n 15, − )( 20 y -20, + x−
12
) + 10, + ( xy
20.
+ 20 1)
+ 1
1 1
3.
is irreducible over Q, p is a prime number.
4 3 2
4. f ( x) = x + x + x + x + 1 is irreducible over Q. Put x = ( y + 1)
=
,
, 5 divide,
Take p = 5 , so
Field Extensions
Definition
4 3 2
= y + 5 y + 10 y + 10 y + 5
is irreducible over Q, hence
Let k be a field. A field K is calld an extension of k if k is subfield of K.
is irreducible over Q.
Let S be a subset of K. k(S) is defined by smallest subfield of k, which contains both k and S.k(S) is an
extension of k. We say k(S) is obtained by adjoining S to k. If
k ( S) : = k( a1, a2 ,........., a n
) .
a finite set, then
If K is an extension of k, then K is a vector space over k. so K has a dimension over k, it may be infinite. The
dimension of K, as a vector space over k is called degree of K over k. Denote it by
dim K k
= degree of K over k
= [ K: k]